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Hi this question has been posed in my integration book where it has been asked to solve it using the given substitution or any other substitution, I've found identical question posted here {1} with different substitutions

The problem is when using the given substitution I end up with integral

$$\int^{\pi/4}_0 \frac{\sin{\theta}}{2x}d\theta $$

Solving for $x$ : $\sqrt{1+x^4} = {(1+x^2)}\cos{\theta}$

According to WolframAlpha: $$x = \frac{|\csc{\theta}||1 \pm \sqrt{\cos{2\theta}}|}{\sqrt{2}} $$

I do know the sign for x is $+ve$ so I'll take $+ve$ solutions after opening the modulus, I do not know which sign to prefer in $ \pm \sqrt{\cos{2\theta}} $ term

Eitherway, I solved for both cases :

$$ \frac{1}{\sqrt{2}}\int^{\pi/4}_0\frac{\sin^2{\theta}}{1 + \sqrt{\cos{2\theta}}} d\theta -(I) $$ $$ \frac{1}{\sqrt{2}}\int^{\pi/4}_0\frac{\sin^2{\theta}}{1 - \sqrt{\cos{2\theta}}} d\theta -(II) $$

Which leads to : $$ \frac{1}{\sqrt{2}}\int^{\pi/4}_0 1 \pm \sqrt{\cos{2\theta}} d\theta = \frac{\pi}{4\sqrt{2}} \pm\frac{\Gamma(\frac{3}{4})^2}{2\sqrt{2}} $$

Here I cheated and used WolframAlpha as I was exhausted.

Both the cases lead to a similar answer with difference of a function unknown to me: $\pm\frac{\Gamma(\frac{3}{4})^2}{2\sqrt{2}}$

The correct answer is :

$$\frac{\pi}{4\sqrt{2}}$$

My main questions are:

  1. What am I doing wrong?
  2. How did author arrive at this ingenious substitution $\sqrt{1+x^4} = {(1+x^2)}\cos{\theta}$ , this doesn't appear to me trial n' error kind of substitution, I have spent hours doing algebraic transformation on this and no matter how you procced and plug the variable your integrand will only have one of these terms only $2x$, $1-x^2$, $1+x^2$ and $ \sqrt{1+x^4}$ and $2x$ being the simplest.

Related question: How do I integrate the following? $\int{\frac{(1+x^{2})\mathrm dx}{(1-x^{2})\sqrt{1+x^{4}}}}$

Integral_spirit
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  • the first link you provided seems to be dead – clathratus Dec 17 '20 at 06:56
  • @clathratus fixed it . I messed up with the markdown :) – Integral_spirit Dec 17 '20 at 07:02
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    Let $J$ be the integral in the title. Using $x\mapsto1/x$, $$J=\int_1^\infty \frac{x^2-1}{x^4+\sqrt{1+x^4}}dx,$$ which seems to be the tiniest bit nicer... – clathratus Dec 17 '20 at 07:03
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    @clathratus just corrected the formatting. term $(1+x^2)$ is in brackets. It will take time for me to get used to the markdown quirks sorry :( – Integral_spirit Dec 17 '20 at 07:06
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    Ah I see. You may find this helpful – clathratus Dec 17 '20 at 07:08
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    $\sin \theta =\frac{x\sqrt{2}}{1+x^2}$. use implicit differentiation to show the integral is equal to $\frac{1}{\sqrt{2}}\int_0^{\pi /4} d\theta $ – Lozenges Dec 17 '20 at 07:41
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    This appears as problem no 42, page 339, A Course of Pure Mathematics, 10th Ed. by G H Hardy. The problem is originally from Mathematical Tripos 1923. – Paramanand Singh Dec 17 '20 at 08:37
  • +1 for comment by @Lozenges and this is easiest way out. There is no need to obtain $x$ in terms of $\theta$. – Paramanand Singh Dec 17 '20 at 08:42
  • @Lozenges using your substitution based on $\sin{\theta}$ gave straightforward answer. I'm curious why the original given substitution doesn't give clear answer $$\sin{\theta}d\theta = \frac{2x(1-x^2)dx}{(1+x^2)\sqrt{1+x^4}}. $$ Plugging $\sin{\theta}$ doesn't clean out other factors.Or am I overlooking some factor? – Integral_spirit Dec 17 '20 at 08:47
  • You should have $(1+x^2)^2\sqrt{1+x^4}$ in denominator in your last comment. Maybe that is where you have an issue. – Paramanand Singh Dec 17 '20 at 09:00
  • @ParamanandSingh while the factor (1+x^2)^2 easily appears with Lozenges substitution . It is somehow escaping my sights when I am doing the cos theta based substitution given in the book.To double check I did the derivative through derivative-calculator website and somehow(1+x^2)^2 'magically' disappears. – Integral_spirit Dec 18 '20 at 13:23
  • I will post an answer based on your substitution which shows all steps. – Paramanand Singh Dec 18 '20 at 13:25
  • Based on my answer did you find your mistake? Let me know if you have further doubt. – Paramanand Singh Dec 18 '20 at 13:47

3 Answers3

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Divide top and bottom by $x^2$:

$$\int_0^1\frac{\frac{1}{x^2}-1}{\left(\frac{1}{x}+x\right)\sqrt{\frac{1}{x^2}+x^2}}dx$$

which suggests using the substitution $t = \frac{1}{x}+x$:

$$\int_2^\infty \frac{dt}{t\sqrt{t^2-2}} = \frac{1}{\sqrt{2}}\sec^{-1}\left(\frac{t}{\sqrt{2}}\right)\Biggr|_2^\infty = \frac{\pi}{4\sqrt{2}}$$

Ninad Munshi
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Let's begin with the substitution suggested in the question. We have $$\cos t=\frac{\sqrt{1+x^4}} {1+x^2}$$ ($\theta$ is replaced by $t$ to reduce burden of typing mathjax). Differentiating we get \begin{align} -\sin t \, dt&=\dfrac {(1+x^2)\cdot \dfrac{4x^3} {2\sqrt{1+x^4}} - 2x\sqrt{1+x^4}} {(1+x^2)^2} \, dx\notag \\ &=\dfrac {(1+x^2)\cdot \dfrac{2x^3} {\sqrt{1+x^4}} - 2x\sqrt{1+x^4}} {(1+x^2)^2} \, dx\notag \\ &= 2x\cdot\dfrac {(1+x^2)\cdot \dfrac{x^2} {\sqrt{1+x^4}} - \sqrt{1+x^4}} {(1+x^2)^2} \, dx\notag \\ &= 2x\dfrac { \dfrac{x^2(1+x^2)-(1+x^4)} {\sqrt{1+x^4}}} {(1+x^2)^2} \, dx\notag \\ &= 2x\dfrac { \dfrac{x^2-1} {\sqrt{1+x^4}}} {(1+x^2)^2} \, dx\notag \\ \implies \sin t \, dt&=\frac{2x(1-x^2)}{(1+x^2)^2\sqrt{1+x^4}}\, dx\notag\\ \end{align} You need to double check your calculations as there is no cancellation in above to reduce the $(1+x^2)^2$ in denominator to $1+x^2$.

Next we note that $$\sin t =\frac{\sqrt {2}x}{1+x^2}$$ and therefore $$dt=\sqrt{2}\cdot\frac{1-x^2}{(1+x^2)\sqrt{1+x^4}}\,dx$$ or $$\frac{1-x^2}{(1+x^2)\sqrt{1+x^4}}\,dx=\frac{dt}{\sqrt{2}}$$ Integrating the above (noting that interval $[0,1]$ for $x$ maps to $[0,\pi/4]$ for $t$) we get $$I=\int_0^1\frac{1-x^2}{(1+x^2)\sqrt{1+x^4}}\,dx=\frac{1}{\sqrt {2}}\int_0^{\pi/4}\,dt=\frac{\pi}{4\sqrt{2}}$$

Paramanand Singh
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  • Thank you for taking out your time to help me out. I' curious how did the author come up with such a clever substitution how did they made relation with cos function inters of 1+x^2 and sqrt(1+x^4). Was there any logic behind construction of such substitution. Has the author(original source) given any explanation regarding this? – Integral_spirit Dec 18 '20 at 13:47
  • @Integral_spirit: as I mentioned in comments this is from an infamous Mathematical Tripos examination conducted in England in 1923. This exam involved tedious questions designed by some group of people who devised many ways to create strange problems. The system of this exam destroyed lot of mathematical creativity and Hardy was so so against it. – Paramanand Singh Dec 18 '20 at 13:50
  • @Integral_spirit: as you can see in another answer there is an alternative and simpler substitution which works so nicely. – Paramanand Singh Dec 18 '20 at 13:52
  • Infamous = (in)famous , so this committee or group of people never published their methods at how they used to arrive at it, just a stroke of luck and brute force? As this is a famous exam I expect people to ask committee of its ways and methods. Strangely, this integral was also not amenable to computer programs. Anyway ,thank you again. – Integral_spirit Dec 18 '20 at 14:15
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Another way. $$I=\int_{0}^{1}\frac{1-x^2}{(1+x^2)\sqrt{1+x^4}}dx \overset{x=-\tan(y/2)}=\\ \frac{\sqrt{2}}{2}\int_{0}^{\pi/2}\frac{\cos{y}}{\sqrt{1+\cos^2{y}}}dy= \frac{\sqrt{2}}{2}\left(\arcsin{\left(\frac{\sqrt{2}\sin{y}}{2}\right)}\right)\Big|_{0}^{\pi/2}=\frac{\sqrt{2}}{2}\cdot\frac{\pi}{4} $$

User
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